Mechanics and Design Finite Element Method: Truss Element SNU School of Mechanical and Aerospace Engineering Engineered system Predicted measure Vehicle Safety Tire Cornering force Cellular phone Reliability Li-ion battery system Thermal behavior Simulation Experiment Applications of FEM in Engineering àThe finite element method (FEM) is a technique for analyzing the behavior of engineered structures subjected to a variety of loads. FEM is the most widely applied computer simulation method in engineering. àThe basic idea is to divide a complicated structure into small and manageable pieces (discretization) and solve the algebraic equation. Introduction
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Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Engineered systemPredicted measure
VehicleSafety
TireCornering force
Cellular phoneReliability
Li-ion battery systemThermal behavior
Simulation
Experiment
Applications of FEM in Engineering
àThe finite element method (FEM) is a technique for analyzing the behavior of engineered structures subjected to a variety of loads. FEM is the most widely applied computer simulation method in engineering.
àThe basic idea is to divide a complicated structure into small and manageable pieces (discretization) and solve the algebraic equation.
Introduction
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
SNU School of Mechanical and Aerospace Engineering
Truss Element in 2D Space
: global coordination system,x y
: local coordination systemˆ ˆ,x y
d : displacement
f : force
u : deformation
L : length
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Most structural analysis problems can be treated as linear static problems, based on the following assumptions.
1. Small deformations (loading pattern is not changed due to the deformed shape)2. Static loads (the load is applied to the structure in slow or steady fashion)3. Elastic materials (no plasticity or failures)
Linear Static Analysis
s e= E e = dudx$$
A Txs = =ddx
AE dudx$$$
FHG
IKJ = 0
Hooke’s Law and Deformation Equation
Equilibrium
Constant
Assumptions1. Truss cannot support shear force. 2. Ignore the effect of lateral deformation
$f y1 0= $f y2 0=
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Direct Stiffness Method
DSM is an approach to calculate a stiffness matrix for a system by directly superposing the stiffness matrixes of all elements. DSM is beneficial to get the stiffness matrix of relatively simple structures consist of several trusses or beams.
Step 1: Determination of element type Step 2: Determination of deformation
$ $u a a x= +1 2 =-F
HGIKJ +
$ $$ $d d
Lx dx x
x2 1
1
$$$u N N dd
x
x
=RST
UVW1 21
2
N xL
N xL1 21= - =
$ $
Assume a linear deformation function as
In vector form
N1, N2: shape function
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Step 3: Strain and stress calculation Step 4: Derivation of element stiffness matrix
Deformation rate - strain
e xx xdu
dxd d
L= =
-$$$ $
2 1
s ex xE=
T A AE d dLx
x x= =-F
HGIKJs
$ $2 1
$ $ $f T AEL
d dx x x1 1 2= - = -e j
$ $ $f T AEL
d dx x x2 2 1= = -e j
$$
$$
ff
AEL
dd
x
x
x
x
1
2
1
2
1 11 1
RSTUVW=
--LNM
OQPRST
UVW
$k AEL
=-
-LNM
OQP
1 11 1
NOTE: In a truss element, stiffness (spring constant, k) is equivalent to AE/L
Stress - strain
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Step 5: Constitution of global stiffness matrix Step 6: Calculation of nodal displacement
K K k e
e
N
= ==å b g
1
F F f e
e
N
= ==ål q ( )
1
Construct a global stiffness matrix in a global coordinate system
• Use boundary conditions • Solve the system of linear algebraic
equations to calculate the nodal deformation in a truss structure
F K d=
Step 7: Calculation of stress and strain in an element
Compute a strain and stress at any point within an element
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Example>
A structure consists of three beams (see below figure). Find (a) the global stiffness
matrix in global coordinate system, (b) the displacements at the node 2 and 3, (c)
the reaction forces at the node 2 and 3. 3000lb loading acts in the positive x-
direction at node 2. The length of the elements is 30in.
Young’s modulus for the elements 1 and 2: E = 30×106 psi
Cross-sectional area for the elements 1 and 2: A = 1in2
Young’s modulus for the elements 3: E = 15×106 psi
Cross-sectional area for the elements 3: A = 2in2
Example (Truss element in 1D space)
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
k k
k
( ) ( )
( )
( )( )
( )( )
1 26
6
1 22 3
36
6
3 4
1 30 1030
1 11 1
101 11 1
2 15 1030
1 11 1
101 11 1
= =´ -
-LNM
OQP =
--LNM
OQP
=´ -
-LNM
OQP =
--LNM
OQP
K
d d d dx x x x
=
-- + -
- + --
L
N
MMMM
O
Q
PPPP10
1 1 0 01 1 1 1 0
0 1 1 1 10 0 1 1
6
1 2 3 4
FFFF
dddd
x
x
x
x
x
x
x
x
1
2
3
4
6
1
2
3
4
10
1 1 0 01 2 1 0
0 1 2 10 0 1 1
RS||
T||
UV||
W||=
-- -
- --
L
N
MMMM
O
Q
PPPP
RS||
T||
UV||
W||
Step 4: Derivation of element stiffness matrix
$k AEL
=-
-LNM
OQP
1 11 1
element #node #
Step 5: Constitution of global stiffness matrix
node #
Answer of (a)
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Step 6: Calculation of nodal displacement
Using the boundary conditions d dx x1 4 0= =
30000
102 11 2
6 2
3
RSTUVW =
--LNM
OQPRST
UVWdd
x
x
So, the displacements for this system are
d in d inx x2 30 002 0 001= =. . . .
Step 7: Calculation of stress and strain in an element
Substituting the displacements to the equation, the load can be obtained as follows.
Answer of (b)
Answer of (c)
xFA
s =
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Selection of Deformation Function
1. The deformation function needs to be continuous within an element.2. The deformation function needs to provide continuity among elements.
3. The deformation function needs to express the displacement of a rigid body and the constant strain in element.
1-D linear deformation function satisfies (1) and (2). à compatibility
a1 considers the motion of a rigid body.considers the constant strain ( )
$ $u a a x= +1 2 : 1-D linear deformation function à completeness
e x du dx a= =$ $ 22 ˆa x
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Truss Element in 2D SpaceVector transformation in 2-D
d i j i j= + = +d d d dx y x y$ $ $ $
Displacement vector d in global coordinate and local coordinate systems.
$$dd
C SS C
dd
x
y
x
y
RS|T|UV|W|=
-LNM
OQPRST
UVWC = cosq S = sinq
C SS C-
LNM
OQP
Relation between two coordination systems
: Transformation Matrix
Global Local
Eq. (1)
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Stiffness matrix in global coordinate system
$$
$$
ff
AEL
dd
x
x
1
2
1
2
1 11 1
RSTUVW=
--LNM
OQPRST
UVW$ $ $f kd=
ffff
k
dddd
x
y
x
y
x
y
x
y
1
1
2
2
1
1
2
2
RS||
T||
UV||
W||=
RS||
T||
UV||
W||
f kd=
The stiffness matrix of truss element in local coordinate system
The stiffness matrix of truss element in global coordinate system
: Step 4 on page 6
: What we want to construct in 2-D space
Let’s calculate the relation between and . $k k
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
$ cos sin
$ cos sin
d d d
d d d
x x y
x x y
1 1 1
2 2 2
= +
= +
q q
q q
$$dd
C SC S
dddd
x
x
x
y
x
y
1
2
1
1
2
2
0 00 0
RSTUVW=LNM
OQPRS||
T||
UV||
W||
$ *d T d= TC S
C S* =
LNM
OQP
0 00 0
$$ff
C SC S
ffff
x
x
x
y
x
y
1
2
1
1
2
2
0 00 0
RSTUVW=LNM
OQPRS||
T||
UV||
W||
$ *f T f=
From Eq.(1),
Simply,
Transform the loading in a same way,
Simply,
Eq. (2)
Eq. (3)
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
$ $ $f kd=
T f kT d* *$=
From Eq.(2) and (3)
We needs the inverse matrix of T*; however, because T* is not the square matrix, it cannot be immediately transformed.
$$$$
dddd
C SS C
C SS C
dddd
x
y
x
y
x
y
x
y
1
1
2
2
1
1
2
2
0 00 0
0 00 0
RS||
T||
UV||
W||=
-
-
L
N
MMMM
O
Q
PPPP
RS||
T||
UV||
W||
$d Td=
TC S
C S* =
LNM
OQP
0 00 0where,
Invite and ( )1 1ˆ ˆ,y yf d ( )2 2
ˆ ˆ,y yf d
In a same way,
f T f=
Transformation Matrix
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
$$$$
$$$$
ffff
AEL
dddd
x
y
x
y
x
y
x
y
1
1
2
2
1
1
2
2
1 0 1 00 0 0 01 0 1 0
0 0 0 0
RS||
T||
UV||
W||=
-
-
L
N
MMMM
O
Q
PPPP
RS||
T||
UV||
W||
Expanded 4×4 matrix
The stiffness matrix of truss elements in global coordinate system
$k
T f k T d= $ f T k T d T k T dT= =-1 $ $
k T k TT= $
T is an orthogonal matrix (TT=T-1).
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
k AEL
C CS C CSS CS S
C CSSymmetry S
=
- -- -
L
N
MMMM
O
Q
PPPP
2 2
2 2
2
2
So,
Assemble the stiffness matrix for the whole system
0 00 0
0 00 0
C SS C
TC SS C
é ùê ú-ê ú=ê úê ú-ë ûk T k TT= $
1 0 1 00 0 0 0ˆ1 0 1 0
0 0 0 0
AEkL
-é ùê úê ú=ê ú-ê úë û
k Ke
e
N( )
=å =
1
f Fe
e
N( )
=å =
1
F K d=
The relation between the nodal loading and displacement in global coordinate system
Eq. (4)
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Example
Three truss elements are assembled and 10,000lb loading is applied at node 1 as shown in
figure. Find the displacements at node 1. For all elements, Young’s modulus:
E psi= ´30 106 , cross-sectional area: A in= 2 2. .
Example (Truss element in 2D space)
Mechanics and Design
Finite Element Method: Truss Element
SNU School of Mechanical and Aerospace Engineering
Step 4: Derivation of element stiffness matrix (from Eq. (4))